Lance Liu

2022-10-25

Laplace Transform: $\frac{1}{{s}^{2}}{(1-{e}^{-s{t}_{0}})}^{2}=\frac{4}{{s}^{2}}{e}^{-s{t}_{0}}{\mathrm{sinh}}^{2}\left(\frac{1}{2}s{t}_{0}\right)$

Verify for the Laplace transform of

$$F(t)=\{\begin{array}{ll}t& 0\le t<{t}_{0}\\ 2{t}_{0}-t& {t}_{0}\le t\le 2{t}_{0}\\ 0& t>2{t}_{0}\end{array}$$

$$\begin{array}{rl}\mathcal{L}\{F(t)\}& ={\int}_{0}^{{t}_{0}}t{e}^{-st}\text{}dt+{\int}_{{t}_{0}}^{2{t}_{0}}(2{t}_{0}-t){e}^{-st}\text{}dt\\ & ={\displaystyle \frac{1}{{s}^{2}}}{(1-{e}^{-s{t}_{0}})}^{2}\\ & ={\displaystyle \frac{4}{{s}^{2}}}{e}^{-s{t}_{0}}{\mathrm{sinh}}^{2}\left({\displaystyle \frac{1}{2}}s{t}_{0}\right)\end{array}$$

I was able to verify everything up to $\frac{1}{{s}^{2}}}{(1-{e}^{-s{t}_{0}})}^{2$, but I don't see how $\frac{1}{{s}^{2}}}{(1-{e}^{-s{t}_{0}})}^{2}={\displaystyle \frac{4}{{s}^{2}}}{e}^{-s{t}_{0}}{\mathrm{sinh}}^{2}\left({\displaystyle \frac{1}{2}}s{t}_{0}\right)$?

Verify for the Laplace transform of

$$F(t)=\{\begin{array}{ll}t& 0\le t<{t}_{0}\\ 2{t}_{0}-t& {t}_{0}\le t\le 2{t}_{0}\\ 0& t>2{t}_{0}\end{array}$$

$$\begin{array}{rl}\mathcal{L}\{F(t)\}& ={\int}_{0}^{{t}_{0}}t{e}^{-st}\text{}dt+{\int}_{{t}_{0}}^{2{t}_{0}}(2{t}_{0}-t){e}^{-st}\text{}dt\\ & ={\displaystyle \frac{1}{{s}^{2}}}{(1-{e}^{-s{t}_{0}})}^{2}\\ & ={\displaystyle \frac{4}{{s}^{2}}}{e}^{-s{t}_{0}}{\mathrm{sinh}}^{2}\left({\displaystyle \frac{1}{2}}s{t}_{0}\right)\end{array}$$

I was able to verify everything up to $\frac{1}{{s}^{2}}}{(1-{e}^{-s{t}_{0}})}^{2$, but I don't see how $\frac{1}{{s}^{2}}}{(1-{e}^{-s{t}_{0}})}^{2}={\displaystyle \frac{4}{{s}^{2}}}{e}^{-s{t}_{0}}{\mathrm{sinh}}^{2}\left({\displaystyle \frac{1}{2}}s{t}_{0}\right)$?

elulamami

Beginner2022-10-26Added 22 answers

$$(1-{e}^{-x}{)}^{2}=4{\textstyle (}\frac{1-{e}^{-x}}{2}{{\textstyle )}}^{2}=4{e}^{-x}{\textstyle (}\frac{{e}^{x/2}-{e}^{-x/2}}{2}{{\textstyle )}}^{2}=4{e}^{-x}{\mathrm{sinh}}^{2}{\textstyle (}\frac{x}{2}{\textstyle )}$$

Here $x=s{t}_{0}$

Here $x=s{t}_{0}$

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$F(s)=\frac{2{s}^{2}+(a-6b)s+{a}^{2}-4ab}{({s}^{2}-{a}^{2})(s-2b)}$

My steps:

$F(s)=\frac{2{s}^{2}+(a-6b)s+{a}^{2}-4ab}{(s+a)(s-a)(s-2b)}$

$=\frac{A}{s+a}+\frac{B}{s-a}+\frac{C}{s-2b}+K$

$K=0$

$A=F(s)\ast (s+a)$